Lexical Analysis - 2

Transcription

1 Lexical Analysis - 2 More regular expressions Finite Automata NFAs and DFAs Scanners JLex - a scanner generator 1

2 Regular Expressions in JLex Symbol - Meaning. Matches a single character (not newline) * Matches 0 or more copies of preceding RE + Matches 1 or more copies of preceding RE? Matches 0 or 1 occurrence of an RE ^ Everything it quotes is matches EXACTLY Matches the beginning of a line $ Matches the end of a line [ ] Character class = matches any character listed; [^ ] implies a match of any character NOT listed ( ) Groups a series of REs into a new RE 2

3 REs in JLex Symbol - Meaning { } Control for repeated matching a specific number of times; a{1,3} means match 1,2, or 3 instances of a \ Used to match a metacharacter or control character; \n matches newline; \* is the character * Means match either the RE proceeding it OR the RE following it RE1/RE2 Means match RE1 but only when followed by RE2; 0/1 will match the 0 in 01 but not in 02 3

4 Exercise Problem: write an RE to match a quoted string such as Hello. Need to decide if a quoted string can go across more than 1 line of text Note: JLex REs are line-input-oriented, unlike formal REs Also, JLex REs make the longest matches possible within a string of characters If multiple REs are given to JLex and several match the same longest expression, the first matching RE is used. 4

5 Finite Automata Automata that recognize strings defined by a regular expression (States, Input symbols, Transitions, Start_state, Set of Final_states) Transitions between states occur on specific input symbols Deterministic automata have only 1 transition per state on a specific input and do not allow transition on the empty string 5

6 Finite Automata Language recognized by automaton is set of strings it accepts by starting in the start state, using transitions corresponding to input symbols in the input string, and processing all input and finishing in a final state RE for integers: A B [ 0-9 ] + Start state Final state 6

7 FAs Nondeterministic finite automaton <{states}, {input symbols} (terminal symbols of a grammar) Transition function ((state,input)--> state), Start state {Final states}> NFA allows more than 1 transition on the same input symbol and/or transitions on ε Deterministic FA allows only 1 transition per input symbol and no ε transitions 7

8 Theoretical results: FAs Set of languages recognizable by NFAs is same as those recognizable by DFAs. There is an algorithm to check for equivalence of two languages recognized by 2 different FAs. RE for reals: ([0-9] + \. [0-9]*) ([0-9] * \. [0-9] + ) Shown: [0-9] +. 0,1,,9 0,1,,9. 0,1,,9 0,1,,9. NFA DFA 8

9 Practical FAs Encode transitions as a table Each column is an input symbol Each row is a state Entry at (s1,i1) is state to transition to when in state s1 and see input i1 Scanner has to try to find longest match in input to a possible token May have to look beyond end of token to do this! 9

10 RE to NFA Conversion Straightforward translation using composition operators of REs For RE ε, s ε f For RE a, terminal symbol, s a f For w,t REs with corresponding NFAs N(w), N(t), w t yields, s ε ε s N(w) s N(t) f f ε ε f 10

11 RE to NFA Conversion For w,t REs with corresponding NFAs N(w), N(t), w t yields, For w RE, w* yields, For (w) RE, use N(w). s s f f N(w) N(t) ε ε ε N(w) f s f ε 11

12 RE to NFA Conversion These drawings follow the Aho, Sethi, Ullman Compiler text and are equivalent to those in Appel For w + use fact that w + = w w* For w? use fact that w? = w ε [abc] = a b c For abc use fact that abc = a b c 12

13 How does an NFA compute? Start off in the start state Compute set S of all states reachable on ε transitions. Given next input symbol is a, calculate set of states T, reachable as transition(s,a) where s S Repeat steps 2,3 until input is exhausted. If final set of states contains a final state, then string has been recognized. 13

14 NFA to DFA Conversion Deterministic computation is desirable if we want a write a scanner as a program Need to convert NFA to equivalent DFA Then can simulate DFA recognition process using tables in program to describe transitions If process ends up in a final state, a token has been recognized 14

15 NFA to DFA Conversion Intuition: whenever there is an ε transition out of a state s, the NFA may go to any of the states reachable in this manner without consuming any input symbols. Call these states the ε-closure of state s. By looking at ε-closures, we form sets of related states in the NFA; these become states in the corresponding DFA Edges in the DFA correspond to sets of edges in the NFA (connecting different ε-closure sets of states) 15

16 NFA to DFA Conversion DFA derived is not the most efficient (smallest possible), but is usually of practical size There are ways of obtaining an optimal DFA by minimizing the numbers of states 16

17 Conversion Algorithm Need two primitive functions ε closure(t), for T a set of states in the NFA Returns a set of NFA states reachable from state s T by ε-transitions move(t, a), for T a set of states in the NFA returns a set of NFA states to which there is a transition on a from some NFA state s T Build set of states (D) and transitions (Dtrans) for the DFA 17

18 Conversion Algorithm Assume all states in NFA are unmarked initially. Let S = ε-closure(start state of NFA). Let D = {S}. while an unmarked state T D do Mark T; input symbols a do { U = ε-closure (move (T, a)); if U D then {add unmarked U to D}; Dtrans(T,a) = U; } endwhile 18

19 Possible Problems Theoretically, for NFA having n states can get DFA with 2 n states, but this doesn t happen in practice. Token is recognized if the ending state of the DFA contains an original final state of the NFA. In case of choice, use final state which represents the earliest rule in the list of productions for tokens 19

20 Optimal DFA There are algorithms for constructing the minimal (smallest) DFA Idea: Assume every state can transition on every input (can create an error state to do this). Try to prove that computation starting at states T and S differs on at least 1 input. If cannot find such an input can merge S and T. Resulting state has the union of their transitions. (ASU, pp141ff) 20